what is solution of y''* y'-y(x)^n=0 ?
wolfram solution is wrong ? There are bugs in many software such as WolframAlpha. Wish wolfram fix the bugs. You can copy equation and paste into MathHand.com, click the ODE button to solve, then click the test button to test its solution.
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On 12/10/2021 10:12 PM, Dr Huang wrote:
what is solution of y''* y'-y(x)^n=0 ?
wolfram solution is wrong ? There are bugs in many software such as WolframAlpha. Wish wolfram fix the bugs. You can copy equation and paste into MathHand.com, click the ODE button to solve, then click the test button to test its solution.
over 500 bugs in
DrHuang.com/index/bugs
This can be solved by transformation on the dependent variable using v(y)=y'(x).
It results in a first order ode in v(y). The solution is then found,
then a new first order ode in y(x) now result which is solved again.
Both are separable.
But the integrals are tricky to do . I used Maple for the integrations.
Here is my solution which Maple's verifies. (there are 3 solutions)
https://12000.org/tmp/121121/foo2.pdf
https://12000.org/tmp/121121/foo2.htm
Maple 2021.2 gives these 3 solutions (Maple can't integrate the separable odes)
Intat(1/(-1/2/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)-1/2*I*3^(1/2)/(1+n)*((3*_a
^(1+n)-_C1)*(1+n)^2)^(1/3)),_a = y(x))-x-_C2 = 0, Intat(1/(-1/2/(1+n)*((3*_a^(1
+n)-_C1)*(1+n)^2)^(1/3)+1/2*I*3^(1/2)/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)),
_a = y(x))-x-_C2 = 0, Intat((1+n)/((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3),_a = y(x))-x
-_C2 = 0
Mathematica 13 can integrate the separable odes' and gives these solutions
in terms of Hypergeometric functions
{{y[x]->InverseFunction[-(((-1/3)^(1/3)*(1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1 *(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1*(1+ #1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((-1)^(2/3)*(1+n)^(1/3) *Hypergeometric2F1[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))] *#1*(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&][x+C[2]]}}
--Nasser
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